这是我的代码。我正在努力检查获胜者。我只是一个初学者，所以请轻松一点。我希望电路板能够改变尺寸。所以，我希望对获胜者的检查可以使用到大小，它不会只检查9个区块。
import java.util.*;
public class TicTacToe {
private String[][] board;
private Scanner console;
public TicTacToe(String[][] table, Scanner console) {
this.board = table;
this.console = console;
}
public void makeTable() {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
board[i][j] = "_";
}
}
}
public void printTable() {
System.out.print(" ");
for (int i = 0; i < board.length; i++) {
System.out.print(" " + i);
}
System.out.println();
for (int i = 0; i < board.length; i++) {
System.out.print(i + "│");
for (int j = 0; j < board[i].length; j++) {
System.out.print(board[i][j] + "│");
}
System.out.println();
}
}
public void play(Scanner console) {
int turn = 0;
String player = "_";
makeTable();
printTable();
while (turn != 9) {
int x = console.nextInt();
int y = console.nextInt();
while (x >= board.length || y >= board[1].length) {
System.out.println("Out of bounce, try again!!!");
x = console.nextInt();
y = console.nextInt();
}
while (board[y][x] != "_") {
System.out.println("Occupied, try again!!!");
x = console.nextInt();
y = console.nextInt();
}
if (turn % 2 == 0) {
player = "X";
} else {
player = "O";
}
board[y][x] = player;
turn++;
printTable();
}
}
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
String[][] board = new String[3][3];
TicTacToe ttt = new TicTacToe(board, console);
ttt.play(console);
}
}
答案 0 :(得分：5)
只有当棋子放在棋盘上时才能获胜，因此您只需要检查涉及刚放在棋盘上的棋子的获胜组合。
例如，如果电路板的当前状态是：
O X O
X X
O
O将他们的作品放在棋盘中间：
O X O
X O X
O
然后你只需要检查涉及这个中间部分的获胜组合，即对角线，以及获胜组合总数（8种组合）中的中间列和中间行（4种组合）。
因此，跟踪最后一步行动对于有效确定董事会是否处于获胜状态至关重要。
答案 1 :(得分：1)
修改
正如一个人已经提到过的，你实际上在做的是检查最后一次移动是否是一个成功的举动。因此，实际上没有必要对每一行，一列和对角线进行强力检查以查看是否存在获胜位置，或者创建某种列表或解决方案表来检查当前的电路板。
你真正需要做的就是检查行，列和对角线（如果移动是在对角线上）进行了移动，并查看是否符合获胜条件。
// Takes the row and column coordinates of the last move made
// and checks to see if that move causes the player to win
public boolean isWinner(int row, int col){
String Player = board[row][col];
int r = row;
int c = col;
boolean onDiagonal = (row == col) || (col == -1 * row + (board.length-1));
boolean HorizontalWin = true, VerticalWin = true;
boolean DiagonalWinOne = true; DiagonalWinTwo = true;
// Check the rows and columns
for(int n = 0; n < board.length; n++){
if(!board[r][n].equals(Player))
HorizontalWin = false;
if(!board[n][c].equals(Player))
VerticalWin = false;
}
// Only check diagonals if the move is on a diagonal
if(onDiagonal){
// Check the diagonals
for(int n = 0; n < board.length; n++){
if(!board[n][n].equals(Player))
DiagonalWinOne = false;
if(!board[n][-1*n+(board.length-1)].equals(Player))
DiagonalWinTwo = false;
}
}
else{
DiagonalWinOne = false;
DiagonalWinTwo = false;
}
boolean hasWon = (HorizontalWin || VerticalWin || DiagonalWinOne || DiagonalWinTwo);
return hasWon;
}
ORIGINAL
有些人已经回答了这个问题，但这只是我的回答。
另外，在你的play方法中，你有一个while循环来检查以确保用户没有指定一个超出范围的移动，但之后你有另一个while循环检查以确保此举是在空旷的地方。您仍然可能需要检查以确保它们的新移动也在边界内，否则您的循环条件将抛出ArrayOutOfBoundsException。
public boolean isWinner(String player){
// Check for N-in-a-row on the rows and columns
for(int i = 0; i < board.length; i++){
boolean verticalWin = true, horizontalWin = true;
for(int j = 0; j < board.length; j++){
if(!board[i][j].equals(player)))
horizontalWin = false;
if(!board[j][i].equals(player))
verticalWin = false;
if(!(horizontalWin || verticalWin))
break;
}
if(horizontalWin || verticalWin)
return true;
}
// If there was a N-in-a-row on the rows or columns
// the method would have returned by now, so we're
// going to check the diagonals
// Check for N-in-a-row on both the diagonals
boolean diagonalWinOne = true, diagonalWinTwo = true;
for(int n = 0; n < board.length; n++){
diagonalWinOne = true;
diagonalWinTwo = true;
int row = board.length - 1 - n;
if(!board[n][n].equals(player))
diagonalWinOne = false;
if(!board[row][n].equals(player))
diagonalWinTwo = false;
if(!(diagonalOne || diagonalTwo))
break;
}
// If either one of the diagonals has N-in-a-row, then there's a winner
if(diagonalWinOne || diagonalWinTwo)
return true;
// Otherwise, no one has won yet
else
return false;
}
答案 2 :(得分：0)
好的，这就是我做Tic-tac-toe时的表现。我使用了String s
String变量，每个玩家一个。 String 现在，这里有神奇的部分，检查胜利者：
6.对于棋盘上的每次点击，开始迭代2d获胜组合。以下是检查某人是否获胜的方法：
String[][] winningCombo = ... initialization ...
for( int i = 0 ; i < winningCombo.length; i++){
for(j = 0; j < winningCombo[i].length; j ++){
char c1 = winningCombo[i][j].charAt(0);
char c2 = winningCombo[i][j].charAt(1);
char c3 = winningCombo[i][j].charAt(2);
if(currentPlayerString.contains(c1) && currentPlayerString.contains(c2) && currentPlayerString.contains(c3)){
// currentPlayer has won if he has all the 3 positions of a winning combo
}
}
}
所以，如果你可以考虑另一种方法，你可以使用它。我将Swing用于用户界面，并使用GridLayout来布局各种JPanel。
答案 3 :(得分：0)
只需检查行，列和两个对角线：
import java.util.Scanner;
public class TTT {
private String[][] board;
private Scanner console;
private int size;
public TTT(String[][] table, Scanner console, int size) {
this.board = table;
this.console = console;
this.size = size;
}
public void makeTable() {
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[i].length; j++) {
board[i][j] = "_";
}
}
}
public void printTable() {
System.out.print(" ");
for (int i = 0; i < board.length; i++) {
System.out.print(" " + i);
}
System.out.println();
for (int i = 0; i < board.length; i++) {
System.out.print(i + "│");
for (int j = 0; j < board[i].length; j++) {
System.out.print(board[i][j] + "│");
}
System.out.println();
}
}
public void play(Scanner console) {
int turn = 0;
String player = "_";
makeTable();
printTable();
while (turn != 9) {
int x = console.nextInt();
int y = console.nextInt();
while (x >= board.length || y >= board[1].length) {
System.out.println("Out of bounce, try again!!!");
x = console.nextInt();
y = console.nextInt();
}
while (board[y][x] != "_") {
System.out.println("Occupied, try again!!!");
x = console.nextInt();
y = console.nextInt();
}
if (turn % 2 == 0) {
player = "X";
} else {
player = "O";
}
board[y][x] = player;
turn++;
printTable();
if(check()){
System.out.println("Player "+player+" won!");
break;
}
}
}
public boolean check(){
//check diagonals
if(check00ToNN()){
return true;
}
if(check0NToN0()){
return true;
}
for(int i = 0 ; i< size ; i++){
if(checkCol(i)){
return true;
}
if(checkRow(i)){
return true;
}
}
return false;
}
public boolean checkRow(int index){
for(int i = 1 ; i< size ; i++){
if(board[i-1][index]!=board[i][index]||board[i][index]=="_"){
return false;
}
}
return true;
}
public boolean checkCol(int index){
for(int i = 1 ; i< size ; i++){
if(board[index][i-1]!=board[index][i]||board[index][i]=="_"){
return false;
}
}
return true;
}
public boolean check00ToNN(){
for(int i = 1 ; i< size ; i++){
if(board[i-1][i-1]!=board[i][i]||board[i][i]=="_"){
return false;
}
}
return true;
}
public boolean check0NToN0(){ //diagonal
for(int i = 1 ; i< size ; i++){
if(board[i-1][size-i-1]!=board[i][size-i]||board[i][size-i]=="_"){
return false;
}
}
return true;
}
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
int size = 3;
String[][] board = new String[size][size];
TTT ttt = new TTT(board, console,size);
ttt.play(console);
}
}
我只是看看是否有胜利者，因为我知道谁有最后一个回合，我知道它是谁。
check()调用真正的checkmethods。
我添加了size，因为它可以扩展。